Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

According to Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.
$$\eqalign{ & \frac{{{p^ \circ } - p}}{{{p^ \circ }}} = \frac{n}{{n + N}}\,\,{\text{or}}\,\,\frac{{\Delta p}}{{{p^ \circ }}} = \frac{n}{{n + N}} \cr & \frac{{10}}{{{p^ \circ }}} = 0.2 \cr & \therefore \,{p^ \circ } = 50\,mm \cr & {\text{For other solution of same solvent}} \cr & \frac{{20}}{{{p^ \circ }}} = \frac{n}{{n + N}}\,\,\,{\text{or}}\,\,\frac{{20}}{{50}} = \frac{n}{{n + N}} \cr & 0.4 = \frac{n}{{n + N}}\,\,{\text{(mole fraction of solute)}} \cr} $$
$$\because $$  Mole fraction of solvent + mole fraction of solute = 1
So, mole fraction of solvent = 1 - 0.4 = 0.6

$${\text{Given, vapour pressure of pure benzene}}$$        $$\left( {{p^ \circ }} \right) = 121.8\,mm$$
$${\text{Vapour pressure of solution}}\,\left( p \right)$$       $$ = 120.2\,mm$$
$$\eqalign{ & {\text{Mass of solute}}\,\left( {{w_B}} \right) = 15\,g \cr & {\text{Mass of benzene}}\left( {{w_A}} \right) = 250\,g \cr & {\text{Molar mass of solvent}}\left( {{m_A}} \right) = 78 \cr & {\text{Molecular weight of solute }}\left( {{m_B}} \right) = ? \cr & {\text{Hence, according to Raoult}}'{\text{s}}\,{\text{law,}} \cr & \frac{{{p^ \circ } - p}}{{{p^ \circ }}} = \frac{{{w_B} \times {m_A}}}{{{m_B} \times {w_A}}} \cr & {p^ \circ } = {\text{vapour pressure of solvent}} \cr & p = {\text{vapour pressure of the solution}} \cr & \frac{{121.8 - 120.2}}{{121.8}} = \frac{{15 \times 78}}{{{m_B} \times 250}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_B} = \frac{{121.8 \times 15 \times 78}}{{250 \times \left( {121.8 - 120.2} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{142506}}{{250 \times 1.6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{142506}}{{400}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 356.26\,g\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Total}}\,\,V.P., \cr & P = P_A^ \circ {X_A} + P_B^ \circ {X_B} = P_A^ \circ {X_A} + P_B^ \circ \left( {1 - {X_A}} \right) \cr & = \left( {P_A^ \circ - P_B^ \circ } \right){X_A} + P_B^ \circ \cr & {\text{Thus,}}\,\,P_B^ \circ = 114\,torr;\,P_A^ \circ - P_B^ \circ = 52 \cr & {\text{or}}\,\,P_A^ \circ = 166\,torr \cr & {\text{Hence}}\,\,P = 166 \times \frac{1}{2} + 114 \times \frac{1}{2} = 140\,torr \cr} $$

$$\eqalign{ & {A_x}{B_y} \rightleftharpoons x{A^{y + }} + y{B^{x - }} \cr & _{{t_{eq}}}^{t = 0}\,\,\,\,\,_{1 - \alpha }^1\,\,\,\,\,_{X\alpha }^0\,\,\,\,\,_{Y\alpha }^0 \cr & {\text{Total no}}{\text{. of moles}}\,(i) = 1 - \alpha + x\alpha + y\alpha \cr & i - 1 = x\alpha + y\alpha - \alpha = \alpha \left( {x + y - 1} \right) \cr & \therefore \,\,\alpha = \frac{{i - 1}}{{\left( {x + y - 1} \right)}} \cr} $$

$$\eqalign{ & {\text{For dilute solution,}} \cr & \frac{{\Delta P}}{{{P^ \circ }}} \Rightarrow \frac{{{n_{{\text{solute}}}}}}{{{n_{{\text{solvent}}}}}}\,\,{\text{For solution in }}A, \cr & \frac{{\Delta {P_A}}}{{P_A^ \circ }} = \frac{{\frac{W}{M}}}{{\frac{{{W_A}}}{{{M_A}}}}} = \frac{W}{M} \times \frac{{{M_A}}}{{{W_A}}}\,\,\,....\left( {\text{i}} \right) \cr & {\text{For solution in}}\,B,\frac{{\Delta {P_B}}}{{P_A^ \circ }} = \frac{W}{M} \times \frac{{{M_B}}}{{{W_B}}}\,\,\,....\left( {{\text{ii}}} \right) \cr & {\text{From (i) and (ii),}} \cr & \frac{{\frac{{\Delta {P_A}}}{{P_A^ \circ }}}}{{\frac{{\Delta {P_B}}}{{P_B^ \circ }}}} = 2 \cr & = \frac{{{M_A}{W_B}}}{{{M_B}{W_A}}} \cr & = \frac{{{M_A}}}{{{M_B}}}\left( {{W_A} = {W_B}} \right) \cr & {M_A} = 2{M_B} \cr} $$

$$Moles$$  of $$2.5\,L$$  of $$1\,M\,NaOH = $$    2.5 × 1 = 2.5
$$Moles$$  of $$3.0\,L$$  of $$0.5\,M\,NaOH = $$    3.0 × 0.5 = 1.5
Total moles of $$NaOH$$  in solution = 2.5 + 1.5 = 4.0
( Total volume of solution = 2.5 + 3.0 = 5.5 $$\,L$$  )
$$\eqalign{ & {\text{Thus,}}\,{M_1} \times {V_1} = {M_2} \times {V_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 = {M_2} \times 5.5 \cr & \therefore \,{\text{Molarity of resultant solution,}} \cr & {M_2} = \frac{4}{{5.5}}M \cr & \,\,\,\,\,\,\,\,\, \simeq 0.73\,M \cr} $$

$$\eqalign{ & \Delta {T_f} = \frac{{{K_f} \times {W_B}}}{{{M_B} \times {W_A}}} \cr & {\text{For cane sugar solution,}} \cr & 2.15\,K = \frac{{{K_f} \times 5}}{{342 \times 0.095}}\left( {\because \,\,95\,g\,\,{\text{of}}\,\,{\text{water}} = 0.095\,\,kg} \right) \cr & {\text{For glucose solution,}} \cr & \Delta {T_f} = \frac{{{K_f} \times 5}}{{180 \times 0.095}} \cr & \frac{{\Delta {T_f}}}{{2.15}} = \frac{{{K_f} \times 5}}{{180 \times 0.095}} \times \frac{{342 \times 0.095}}{{{K_f} \times 5}} \cr & \Delta {T_f} = \frac{{342}}{{180}} \times 2.15 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 4.085\,K \cr & {\text{Freezing point of glucose solution}} \cr & = 273.15 - 4.085 \cr & = 269.07\,K \cr} $$

\[\underset{\begin{smallmatrix} \text{Initially} \\ \\ \text{At equil} \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} C \\ C\left( 1-\alpha \right) \end{smallmatrix}}{\mathop{2{{C}_{6}}{{H}_{5}}OH}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ C\alpha /2 \end{smallmatrix}}{\mathop{{{\left( {{C}_{6}}{{H}_{5}}OH \right)}_{2}}}}\,\]

$$\eqalign{ & {\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{moles}} \cr & = C - C\alpha + \left( {\frac{{C\alpha }}{2}} \right) \cr & = C - \frac{{C\alpha }}{2} \cr & = \frac{{2C - C\alpha }}{2} \cr & = \frac{{C\left( {2 - \alpha } \right)}}{2} \cr & C = \frac{{{\text{Weight}}}}{{{\text{Molecular}}\,{\text{weight}}}} \cr & \,\,\,\, = \frac{{75.2}}{{94}} \cr & \,\,\,\, = 0.8 \cr & \Delta {T_f} = {K_f} \times m \cr & \Rightarrow 7 = 14 \times 0.8\left( {\frac{{2 - \alpha }}{2}} \right)\,\,\,{\text{or}},\,\,1 = 0.8\left( {2 - \alpha } \right) \cr & \alpha = 0.75 = 75\% \cr} $$

NOTE: On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases.

Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solute-solvent interactions. In such solution, the interactions among molecules becomes weaker. Therefore their escaping tendency increases which results in the increase in their partial vapour pressures. In pure methanol there exists intermolecular $$H$$–bonding.

On adding benzene, its molecules come between methanol molecules there by breaking $$H$$-bonds which weaken intermolecular forces. This results in increase in vapour pressure.